A particle is projected from the bottom of an inclined plane of inclination  300 with velocity of 40 m/s at an angle of 600 with horizontal. Then  (Take g=10 m/s2 )

Velocity vector will be parallel to the plane when component of velocity Perpendicular to the plane is zero. Here  α=600,β=300  Velocity vector will be parallel to the plane when component of velocity Perpendicular to the plane is zeroVy=uy+ayt=0vy=usin⁡(α−β)−(gcos⁡β)t=0   In this problem, β=30∘,α=60∘ and u=40msecvy=40×sin⁡30∘−10cos⁡30∘=0⇒20=53t⇒t=43sec At this instant the particle will have vx component only vx=ux+axt  vx=40cos⁡30∘−10×sin⁡30∘×43vx=403m/sec Speed of the particle  =403m/s